Oops I broke the laws of math.

[ninjaKIWI]

Originally Posted by bLasTamos

The sum S is infinite, it doesn't make any sesne to say S-A is 4S because nothing can be the 4 times more infinite than something else.

Well, there can be different sizes of infinity, can't there?

Like, an infinity that grows 2,4,6,8 would be considered 'larger' than one that grows 1,2,3,4.

[emerald000]

Originally Posted by ninjaKIWI

Well, there can be different sizes of infinity, can't there? Like, an infinity that grows 2,4,6,8 would be considered 'larger' than one that grows 1,2,3,4.

Yes and no. Yes, there are different "sizes" of infinity. The size of an infinite group is called its cardinality.

But those two ensembles have the same cardinality: aleph-null. You can see that by pairing each term. 1st term with 1st term, etc. You won't be able to find a number in one ensemble that doesn't have a pairing. If you want a larger infinity, you have to look in non-countable ensembles. Real numbers, for example. There are infinitely more real numbers than there are integers.

[Niala]

Don't know if this has been posted yet, but:

0.999999999999etc.=1
x=0.99999
Therefore 10x=9.99999
10x-x=9.99999-0.99999=
9x=9
9x/9=9/9
x=1

o0o0o0o0o

[x After Dawn x]

Originally Posted by Niala

Don't know if this has been posted yet, but:

0.999999999999etc.=1
x=0.99999
Therefore 10x=9.99999
10x-x=9.99999-0.99999=
9x=9
9x/9=9/9
x=1

o0o0o0o0o

yes, except this one is actually true

it's already been proven through many proofs that 0.9... = 1

another one could be 1/3 = 0.3..., 2/3 = 0.6..., then 3/3 = 0.9..., or 1

also 1/9 = 0.1..., 2/9 = 0.2..., 5/9 = 0.5..., etc. And if you look, 9/9 should equal 0.9... according to that pattern but it's also 1; you could also say 8/9 (0.8...) + 1/9 (0.1...) = 0.9..., or 1.

[stargroup]

you could also say 2/9 + 7/9 = 1

you could also say 4/9 + 5/9 = 1

can we not point out obvious solutions thanks

[Ghakimx]

Proof 6 = Vote4Nixon

[MrRubix]

Building on emerald's logic, let me pose to you all a question.


Consider Pascal's Triangle (each number is the sum of the two numbers above it):

http://upload.wikimedia.org/wikipedi...triangle_5.svg

I shall make the claim that given a Pascal's Triangle that is infinitely large, I can throw a dart at this triangle -- at ANY spot -- and have a 0 percent change of hitting an odd number. Not "roughly zero" -- EXACTLY zero. Not "it's very unlikely," but rather "literally impossible."


Mathematical truth, or is Rubix trying to pull a fast one on you all?

[leonid]

I threw one and it missed the triangle

[-zeroSKILL-]

i'd like to see how the hell you would prove it. but i am curious

[MrRubix]

Originally Posted by leonid

I threw one and it missed the triangle

That takes some srs skill

[emerald000]

That has to be about combinatorics, but I can't see how you would do that. Heck, I still have to see that's the right answer.

EDIT: Just saw a beautiful pattern. I see now that it has a zero chance of getting an odd number. Now to prove it...

[Ghakimx]

Yay, zero percentage of getting an odd number.

[Relambrien]

Originally Posted by MrRubix

Building on emerald's logic, let me pose to you all a question.


Consider Pascal's Triangle (each number is the sum of the two numbers above it):

http://upload.wikimedia.org/wikipedi...triangle_5.svg

I shall make the claim that given a Pascal's Triangle that is infinitely large, I can throw a dart at this triangle -- at ANY spot -- and have a 0 percent change of hitting an odd number. Not "roughly zero" -- EXACTLY zero. Not "it's very unlikely," but rather "literally impossible."


Mathematical truth, or is Rubix trying to pull a fast one on you all?

30-second logic:

In a practical sense, because there are odd numbers in the triangle, even an infinitely large one, there will always be points which the dart can hit which will be odd numbers. Specifically the first and last numbers of each row.

In a strictly mathematical probability sense, as the size of the triangle increases, the ratio of evens : odds rises (note: I'm not entirely sure of this, actually). So as the size of the triangle approaches infinity, the probability of any arbitrary number on the triangle being odd approaches zero.

disclaimer no this probably isn't the correct reasoning blah blah blah it's just a result of quick look

edit: I should get my discrete math professor to look at this (John Mackey, best math teacher I've ever had)

[-zeroSKILL-]

i get it now... but the way you prove it through philosophy is BS >.>

edit:
ninja'd... I was not referring to above posters logic.

[emerald000]

Well, I guess it could simply be proven by saying that the numbers modulo 2 makes a wonderful Sierpinski triangle (slightly cut off if the number of rows isn't a power of 2). The odd numbers are the plotted points. So since the area of the Sierpinski triangle after an infinity of iterations is null, (each iteration removes 1/4 of the remaining area-> converges towards zero) the probability of getting an odd number is null.

EDIT: Relambrian, the hard part is actually proving the ratio evendd is bigger than one.

EDIT2: After rereading, I see it is not really clear, but I understand myself and that's what is important. Right... Right?

[leonid]

Want to prove by induction: If there are k odd numbers in the first 2^(n-1) rows of the triangle, there are 3k odd numbers in the first 2^n rows of the triangle, having odd number in every cell of the 2^nth row.

Base case: n=1: 1st row has one odd number (1), and the first two rows contain three odd numbers (three 1's), and every cell of the 2nd row is odd.

Assume the theorem works for n >= 1. Now to prove this works for n+1.

Since every cell of the 2^nth row is odd, every cell except for the first and the last one in the (2^n+1)th row must be even.

Thus, the (2^n+2)th row must have odd numbers at 1st, 2nd, (2^n+1)th, and (2^n+2)th cell, and all remaining cells must contain even numbers,
according to the definition of pascal triangle.

Likewise, (2^n+m)th row must contain odd number at mth cell from left and from right, and even numbers inbetween, for every 1 <= m <= 2^n

This makes each half of the rows from (2^n+1)th to (2^(n+1))th contain triangle, whose borders consist of odd numbers. Thus we can apply induction hypothesis and conclude that each triangle contains the same number of odd numbers as the first 2^n rows do.

Hence the first 2^(n+1) rows contain three times the number of odd numbers in the first 2^n rows.

Also, because of induction hypothesis, each triangle's bottom must only consist of odd numbers, making the whole 2^(n+1)th row all odd.


Therefore we proved that the first 2^n rows contain three times the number of odd numbers in the first 2^(n-1) rows,
which implies that the first 2^n rows contain 3^n odd numbers.

So, the chance for hitting an odd number when throwing a dart toward the triangle of height 2^n is

(3^n) / [(2^n)(2^n+1)/2]

= 2(3^n) / (2^n)(2^n+1)
= 2(1.5^n) / (2^n+1) < 2(1.5^n) / (2^n) = 2(0.75^n)

As n goes to infinity, the chance to hit the odd number will converge to zero.

[leonid]

I wasted 20 minutes of my life

[MrRubix]

Yup yup, good job, guys. That's really all there is to it -- but imo it is a great problem for understanding exactly what infinity means.


Also, for those that give you **** for the .999... argument, consider this:

1/3 (0.333...) in base 10 is .1 in base 3, and 1/10 in base 10 is 0.0022... in base 3.

[Ghakimx]

I just add each row.

[leonid]

Rubix, your avatar looks similar to shadow of pedobear